# How do you rationalize the denominator sqrt2/(2-sqrt2)?

Apr 1, 2015

Multiply by $1$ in the form: $\frac{2 + \sqrt{2}}{2 + \sqrt{2}}$.
(This doesn't change the value of the number, it only changes how the number is written.)

$\frac{\sqrt{2}}{2 - \sqrt{2}} = \frac{\sqrt{2}}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2} \left(2 + \sqrt{2}\right)}{\left(2 - \sqrt{2}\right) \left(2 + \sqrt{2}\right)}$. So,

$\frac{\sqrt{2}}{2 - \sqrt{2}} = \frac{2 \sqrt{2} + {\left(\sqrt{2}\right)}^{2}}{4 - {\left(\sqrt{2}\right)}^{2}} = \frac{2 \sqrt{2} + 2}{4 - 2}$

$\frac{\sqrt{2}}{2 - \sqrt{2}} = \frac{2 \sqrt{2} + 2}{2} = \frac{2 \left(1 + \sqrt{2}\right)}{2} = 1 + \sqrt{2}$

Why does it work?

We can verify that $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$ for any numbers, $a$ and $b$.

So we can also verify that $\left(a - \sqrt{c}\right) \left(a + \sqrt{c}\right) = {a}^{2} - c$. (no square roots).

By the way, we can also verify that $\left(\sqrt{a} - \sqrt{b}\right) \left(\sqrt{a} + \sqrt{b}\right)$ has no square roots.